What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP
2 answers:
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Answer: maximum length of the nanowire is 510 nm
Explanation:
From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K
Thermal conductivity of silicon carbide k = 30 W/m.K
Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m
lets consider the equation for the value of m
m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )
m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04
now lets find the value of h/mk
h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353
lets consider the value θ/θb by using the equation
θ/θb = (T - T∞) / (T - T∞)
θ/θb = (3000 - 8000) / (2400 - 8000)
= 0.893
the temperature distribution at steady-state is expressed as;
θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]
so we substitute
0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]
L = 510 × 10⁻⁹m
L = 510 nm
therefore maximum length of the nanowire is 510 nm
Answer:
Q=0.95 W/m
Explanation:
Given that
Outer diameter = 0.3 m
Thermal conductivity of material
So the mean conductivity
So heat conduction through cylinder
Q=0.95 W/m